How do you find the exact value of ${log}_{2} (- 16)$ $log_2 (-16)$ ?

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How do you find the exact value of ${log}_{2} (- 16)$ $log_2 (-16)$ ?

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Answer 1 · 2017-01-26T07:03:40

$= 4 + i \frac{(2 n + 1) π}{ln 2}, n = 0, \pm 1, \pm 2, \pm 3, . .$ $=4+i((2n+1)pi)/ln2, n = 0, +-1, +-2, +-3, ..$ , where $i = \sqrt{- 1}$ $i =sqrt(-1)$

Explanation:

$W h e n x \leq 0, {log}_{b} x$ $When x <=0, log_b x$ has values that are complex.

Here,

${log}_{2} (- 16)$ $log_2(-16)$

$= \frac{ln (- 16)}{ln 2}$ $=ln (-16)/ln 2$

$= \frac{ln ({(\pm i)}^{2} 2^{4})}{ln 2}$ $=ln ((+-i)^2 2^4)/ln 2$

$= \frac{2 ln (\pm i) + 4 ln 2}{ln 2}$ $= (2 ln (+-i) + 4ln 2)/ln2$

$= 4 + \frac{2}{ln 2} {ln e}^{i (2 n + 1) \frac{π}{2}}, n = 0, \pm 1, \pm 2, \pm 3, . .$ $=4+2/ln2 lne^(i(2n+1)pi/2), n = 0, +-1, +-2, +-3, ..$

$= 4 + i \frac{(2 n + 1) π}{ln 2}, n = 0, \pm 1, \pm 2, \pm 3, . .$ $=4+i((2n+1)pi)/ln2, n = 0, +-1, +-2, +-3, ..$

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How do you find the exact value of ${log}_{2} (- 16)$ $log_2 (-16)$ ?

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How do you find the exact value of ${log}_{2} (- 16)$ $log_2 (-16)$ ?